MCQ
$\int {\left( {\sin x\cos x\cos 2x\cos 4x\cos 8x} \right)dx}$ equal
- A$\frac{{ - 1}}{{128}}\cos 16x + C$
- B$\frac{{1}}{{256}}\cos 16x + C$
- C$\frac{{ - 1}}{{256}}\sin 16x + C$
- ✓$\frac{{ - 1}}{{256}}\cos 16x + C$
$ = \frac{1}{4}\int {\sin } \,4x\cos \,4x\cos \,8x\,dx$
$ = \frac{1}{8}\int {\sin } \,8x\cos \,8x\,dx = \frac{1}{{16}}\int {\sin } \,16x\,dx$
$=\frac{-1}{256} \cos 16 x+C$
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$\left|\begin{array}{ccc}\cos ^{2} x & 1+\sin ^{2} x & \sin 2 x \\ 1+\cos ^{2} x & \sin ^{2} x & \sin 2 x \\ \cos ^{2} x & \sin ^{2} x & 1+\sin 2 x\end{array}\right|$.
Then the ordered pair $( m , M )$ is equal to
$\alpha=\sum_{ k =1}^{\infty} \sin ^{2 k}\left(\frac{\pi}{6}\right)$
Let $g:[0,1] \rightarrow R$ be the function defined by
$g( x )=2^{\alpha x }+2^{\alpha(1- x )}$
Then, which of the following statements is/are $TRUE$?
$(A)$ The minimum value of $g( x )$ is $2^{\frac{7}{6}}$
$(B)$ The maximum value of $g( x )$ is $1+2^{\frac{1}{3}}$
$(C)$ The function $g( x )$ attains its maximum at more than one point
$(D)$ The function $g( x )$ attains its minimum at more than one point